Let there be the function f(alpha)= (L( v_a - v_0 sin alpha))/(v_0 cos alpha) Where L ,v_a and v_o are constants. Determine alpha such that f(alpha) is minimal ?

2 Answers
Jul 5, 2018

f(alpha)= (L( v_a - v_0 sin alpha))/(v_0 cos alpha)

= L( v_a/v_o sec alpha - tanalpha)

L is just a scaling factor so it can be removed, thus writing:

  • F(alpha) = f(alpha)/L qquad F'(alpha) = (f'(alpha))/L qquad " etc "

To optimise F(alpha), differentiate wrt alpha:

  • F_alpha = v_a/v_o secalphatanalpha - sec^2alpha = 0

implies underbrace(sec^2alpha)_(gt 0, forall alpha)( v_a/v_o sinalpha - 1 ) qquad = square qquad = 0

Critical points can, therefore, only occur under these rules:

  • sinalpha = v_o/v_a qquad qquad implies {(v_o lt= v_a),(cos alpha = sqrt(v_a^2 - v_o^2)/v_a),(tanalpha = v_o/sqrt(v_a^2 - v_o^2) ):}

To determine the max or min nature of the Critical Point, grab the second derivative of square, using the product rule:

F_(alpha alpha) = (2 sec^2alpha tanalpha )( underbrace(v_a/v_o sinalpha - 1)_(= 0) ) + sec^2alpha(v_a/v_o cos alpha)

= sec^2 alpha (v_a/v_o cos alpha) = v_a/v_o sec alpha

:. F_(alpha alpha) = color(blue)(v_a/v_o) * 1/color(green)(sqrt(1 - v_o^2/v_a^2))

With assumption v_o lt= v_a, then the green term is positive; whole expression is positive provided v_a, v_o gt 0.

Accordingly, there is a min when this occurs:

  • sinalpha = v_o/v_a

This min will be periodic.

I have solved this way. See the answer below:
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